a wall is 15 ft high and 10 ft from a house. find the length x of the shortest ladder which can go from the…

a wall is 15 ft high and 10 ft from a house. find the length x of the shortest ladder which can go from the top of the wall to a window 20.5 ft above the ground. x = ? ft. round to the nearest hundredth.
Answer
Explanation:
Step1: Set up the proportion
Let the angle of the ladder with the ground be (\theta). The length of the ladder (x) can be expressed as (x=\frac{15}{\sin\theta}+\frac{10}{\cos\theta}).
Step2: Differentiate the function
Differentiate (x) with respect to (\theta): (x^\prime=-\frac{15\cos\theta}{\sin^{2}\theta}+\frac{10\sin\theta}{\cos^{2}\theta}) Set (x^\prime = 0) for critical points: (-\frac{15\cos\theta}{\sin^{2}\theta}+\frac{10\sin\theta}{\cos^{2}\theta}=0) (\frac{15\cos\theta}{\sin^{2}\theta}=\frac{10\sin\theta}{\cos^{2}\theta}) (15\cos^{3}\theta = 10\sin^{3}\theta) (\tan^{3}\theta=\frac{15}{10}=\frac{3}{2}) (\tan\theta=\sqrt[3]{\frac{3}{2}})
Step3: Calculate (\sin\theta) and (\cos\theta)
If (\tan\theta=\sqrt[3]{\frac{3}{2}}), then (\sin\theta=\frac{\sqrt[3]{\frac{3}{2}}}{\sqrt{1 + (\sqrt[3]{\frac{3}{2}})^{2}}}) and (\cos\theta=\frac{1}{\sqrt{1+(\sqrt[3]{\frac{3}{2}})^{2}}}) Another way: Let's use similar - triangles. Let the total height from the ground to the window (h = 20.5) ft and the height of the wall (a = 15) ft. The distance between the wall and the house (b = 10) ft. By similar - triangles, if we assume the length of the part of the ladder above the wall is (y) and the length of the part of the ladder from the ground to the wall is (z). We have (\frac{y}{10}=\frac{h - 15}{z}) and (x=y + z). Also, from similar - triangles (\frac{15}{z}=\frac{20.5}{x}) (using the property of similar right - triangles). Let's use the formula (x=\sqrt{(10+\frac{15\times10}{h - 15})^{2}+h^{2}}) (derived from similar - triangles and Pythagorean theorem). Substitute (h = 20.5) (x=\sqrt{(10+\frac{15\times10}{20.5 - 15})^{2}+20.5^{2}}) First, calculate (\frac{15\times10}{20.5 - 15}=\frac{150}{5.5}=\frac{300}{11}\approx27.27) (10+\frac{300}{11}=\frac{110 + 300}{11}=\frac{410}{11}\approx37.27) (x=\sqrt{(\frac{410}{11})^{2}+20.5^{2}}=\sqrt{\frac{168100}{121}+420.25}=\sqrt{\frac{168100+420.25\times121}{121}}=\sqrt{\frac{168100 + 50840.25}{121}}=\sqrt{\frac{218940.25}{121}}\approx\sqrt{1810.25}\approx42.55)
Answer:
(42.55) ft